Different size of fixed array (c++)

Asked Jan 19 '18 at 14:39

Active Jan 19 '18 at 14:39

Viewed 54 times

I want to create a programm, that counts how much time one array is contained in another one.

#include <iostream>
#include <vector>
#include <string>
#include <stdexcept>
#include <algorithm>
#include <cmath>
#include <list>
using namespace std;

int func(char one[],char two[]){
    int result = 0;
    int length_one = sizeof(one)/sizeof(*one) - 1;
    int length_two = sizeof(two)/sizeof(*two) - 1;
    for(int i = 0;i<length_one;++i){
        int temp = 0;
        if(one[i] != two[0]) continue;
        else{
            for(int j = 0;j<length_two;++j){
                if(two[j] != one[j+i])break;
                else{
                    temp+=1;
                }
            }
            if(temp == length_two){
                result +=1;
            }
        }

    }
    return result;
}


int main(){
    char one[] = "Hello everybody";
    char two[]= "every";
    cout << func(one,two);

}

I checked separately the line: int length_two = sizeof(two)/sizeof(*two) - 1; the size is equal to 5. However in my code int length_two is equal to 7. How is it possible?

asked Jan 19 '18 at 14:39

Rob_Fir

well, if you try `int x=sizeof(two), y=sizeof(*two);` you'd see – UKMonkey Jan 19 '18 at 14:41
@UKMonkey I saw it. When I test it separatly, it gives me 6/1 - 1 = 5(as it should be. But in this programm the result is 7. How? – Rob_Fir Jan 19 '18 at 14:45
prints 7 for me. I'd go with you typed + rather than - – UKMonkey Jan 19 '18 at 14:47
1

I'd go with OP tested that in main and hasn't read the dupe. – Baum mit Augen Jan 19 '18 at 14:48
@BaummitAugen ah, I see. How should I correct it, so that size in function become 5 instead of 7? – Rob_Fir Jan 19 '18 at 14:59
https://stackoverflow.com/a/1975790/3002139 – Baum mit Augen Jan 19 '18 at 16:19

Different size of fixed array (c++)

0 Answers0